# Prove that $f$ is locally Lipschitz-continuous [duplicate]

Let $f:E\subset\mathbb{R^n}\rightarrow\mathbb{R^n}$ be a $C^1$ function, where $E$ is open. Show that $f$ is locally Lipschitz-continuous, that is, for all compact $K\subset E$ we have $f$ restrict to $K$ Lipschitz-continuous.

I try to take a open ball in $E$ to force convexity, and use a auxiliary function $g: [0,1] \rightarrow \mathbb{R^n}$ such that $g(t) = f((1-t)x+ty)$, but this don't work.

## marked as duplicate by John B, JonMark Perry, Community♦May 29 '16 at 23:39

I'll do the $\mathbb{R}$ case. The reasoning can be generalized to $\mathbb{R}^n$. Let $x\in E$ and $\epsilon>0$ small enough, then as $f$ is $C^1$, there exists a neighborhood of $x$ (we can take a sphere $B(x,r)$ of some radius $r>0$) such that $$f'(x')\in[f'(x)-\epsilon,f'(x)+\epsilon].$$ Therefore, for any two points $y,z\in B(x,r)$ we have $$|f(y)-f(z)| = \left|\int_y^zf'(s)ds\right|\le|z-y|(|f'(x)|+\epsilon)$$ (for the general case, integrate over the straight line relying $y$ and $z$).